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How to track a flight that I am flying in using the GPS antenna in my mobile phone (without internet)?
I am in India. I booked my first flight ticket a few weeks ago and am very interested in identifying locations as well as geographical and human structures that I will see.
I have downloaded many GB's of offline maps on Google Maps. I had hoped that I will be able to use the GPS antenna on my mobile phone to get my latitude and longitude, and the Google Maps app would mark that location on the pre-downloaded offline map.
However, it seems that this doesn't work. I requested a friend to test it on his last flight.
I have two questions:

Why doesn't this work?

Is there a way I can track the plane without internet using the GPS antenna in my mobile phone?

Would Wi-Fi and a flight tracking site accomplish your purpose?

In what way doesn't it work? If it is a problem with GPS signal reception, quiet flyer's suggestions make sense. If it is a problem with the Google Maps app not liking to work offline, you might try a more dedicated GPS app such as Commander Compass (there is a free version).

@randomhead Thank you! I will try this app, as well as quiet flyer's recommendations.

If you have a window seat and enough GPS signal, you might try using a better app than Google maps.  There are various free Android apps capable of using offline mapping for hiking etc.  The one I use (mainly a bike computer app)  isn't free after a distance limit, so I won't recommend it here, and my other experience is out of date (hence no answer).  OSMAnd might be worth a shot, for example, or Navit.  Test it in a car first

Duplicate on Travel Stack Exchange: GPS when flying over the ocean

GPS reception inside an airliner fuselage will be poor.  Holding the phone right up to the window will help some.  I've often used a handheld Garmin Etrex GPS in exactly this way.
However some airlines prohibit use of such electronic devices.  If you hold a GPS device right next to a radio antenna you can in fact pick up some electronic noise, so there's a concern that there may be interference with aircraft navigation instruments.
And it goes without saying that if you are going to be using your device in "airplane" mode, then you should confirm, on the ground, that your device does in fact receive and use GPS signals when in "airplane" mode.   A quick google search suggests that this normally is the case, despite the potential issue noted above.
One problem is that if you are instructed to turn the device all the way off before takeoff, then when you turn the device on later in the flight, it has to calculate its position from an incorrect assumed starting point, while moving at high speed.  This can be a challenge and may take quite some time.  Try not to lose the satellite lock, once you've finally acquired it.
Here are three "quora" links that include some answers from people who successfully did what you are proposing -- https://www.quora.com/Does-mobile-GPS-work-onboard-airplanes , https://www.quora.com/Why-does-GPS-work-in-airplane-mode-and-in-the-mountains-with-no-cell-tower-nearby-but-doesn-t-work-on-planes , https://www.quora.com/Why-is-it-not-possible-to-use-the-Google-Maps-satellite-GPS-when-inside-a-flying-plane

I used Waze to track a flight once. It did register that I was going about 500MPH (which seemed reasonable in the air). However, the speed did jump around considerably, and my track was erratic, leading me to believe that my GPS reception in the plane was sketchy at best. I was in the middle seat with my kid staring out the window.

My plane has a bubble canopy with no metal above, so my phone's GPS works fine, and I have a program called Naviator on my phone that I use for my VFR flying. I haven't used the airplane's nav radio or even the on-board Garmin GPS 295 in ages.  The phone should work well in an airliner if held right against the window.

Or you can just use in-flight wifi and any number of online flight trackers (just realize that position may have some delay).

@JohnK The problem is an airliner window will only allow seeing satellites in one direction, so you may not see enough to get a lock or, due to triangulation error, it will be so inaccurate as to be useless.


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Forces in a slipping turn
I have trouble understanding why the ball (on the turn & slip indicator) falls into the turn e.g. drop to the left when aircraft is rolled left, when in an uncoordinated turn (slipping turn). What are the forces that goes into this?
From reading up online, many sources quote that in a slipping turn, the aircraft is banked too much for the rate of turn. So, the horizontal lift component is greater than centrifugal force. As a result, the ball drops into the turn since there is less g force to "centralise" the ball.
Isn't centrifugal force suppose to be equal and opposite of the horizontal component of lift. How can being in a slipping turn make the centrifugal force lesser? Essentially, what is so special about slipping turn can allow the centrifugal force to change?

Image from: http://avstop.com/ac/flighttrainghandbook/forcesinturns.html

Thank you for your replies. As much as I agree that the links you shared have their merits, I am still unable to pinpoint how exactly a slipping turn can have an effect on the centrifugal force.

I fail to understand how this answer does not answer your question. You may elaborate that.

Please pardon me but I have trouble relating that particular answer to my question. I am under the impression that the centrifugal force generated is EQUAL and opposite to the horizontal component of lift. As such,  I cannot relate to why the centrifugal force can be changed.

The answers in the linked questions include all acceleration, including the one that you can model with a centrifugal force (which is only here to take into account the fact that when turning, the aircraft is not an inertial frame of reference)

I guess my lack of comprehension is a reflection of my weak fundamentals in physics. Perhaps this is beyond me right now. Thank you so much for your replies.

thus you can divide your question in smaller subproblems and ask them in separate questions, including on physics.SE

@Manu H many of these "answers" only serve to confuse a very simple concept.

This may be another case where the new question is better (at least simpler and easier for a beginner to understand) than the ones it is potentially duplicating-- as discussed on ASE Meta here -- https://aviation.meta.stackexchange.com/questions/4092/duplicates-should-we-close-the-new-question-if-its-better-than-the-old-one

"many of these "answers" only serve to confuse a very simple concept."  YES!!!  I only wish I would have been logged on in time to vote to close this earlier.  Good grief already...

So glad I logged on in time to save this poor soul needless confusion.  It's a hard thing to be adrift, lost at sea, with only the official FAA explanation for guidance, as mirrored in countless flight training materials...

Shame that the people closing this question seem "adrift at sea" too.  @quiet flyer once asked "what is missing in these diagrams", I would submit: a picture of the turn itself.  the yaw component W.R.T. earth is what causes the ball to move outside the circle.  This is why we "create a bank", just like a roadbuilder, to center the belly of the plane on the new combined G vector.  In a shallow turn, the rudder does the yawing W.R.T. earth.  "bank and yankers" know, in a steep turn it's the elevator too.

@RobertDiGiovanni -- too bad you have issues w/ chat, we could discuss more there --

"Essentially, what is so special about slipping turn can allow the centrifugal force to change?" -- this is very concise and very well stated. This is not the main thrust of any of the questions that this answer has been suggested to be a duplicate of. In my opinion this sentence alone is enough to qualify this question as not a duplicate of some other ASE question.  Please give the matter some thought and consider casting a vote to re-open, or if your reputation score doesn't permit this, flag for re-opening with a note as to why.

Continuing along the same vein-- I looked at the questions suggested as possible duplicates- one is about the effect of crosswinds which is completely irrelevant here. One is about what the slip-skid ball measures. Even if one completely understood what the slip-skid ball measures, which can be said to be the balance between centrifugal force and weight, it would not answer the question "Essentially, what is so special about slipping turn can allow the centrifugal force to change?". Just because some of the answers overlap, does not mean the questions do. Please consider, and vote accordingly.

Re -- "I fail to understand how this answer does not answer your question. You may elaborate that." -- see immediately preceding comment. Just because some of the answers overlap, does not mean that the actual questions are sufficiently the same to be considered to be duplicates.  Just because the answer to a newer question may be found in the answer to an older question, does not mean that the two questions are duplicates.  Many answers are broader in scope than the bare minimum content that would be needed to answer the actual question that was asked.  Please consider, and vote to re-open.

@quietFlyer,  Didn't see this until today...  But simplest answer is just that, simple. As the only forces that can be felt are real forces, not fictitious, the only force that affects the ball is the aerodynamic force on the aircraft. SO the ball sits to one side of the race ONLY when the total Vector summation of all aerodynamic forces on the aircraft is aligned at an angle which is off to one side of the vertical "z" axis of the aircraft  (not perpendicular to the wings). it lays at a point exactly opposite of the angle of the total Aerodynamic force vector. That's it. Nothing else matters.

I have trouble understanding why the ball (on the turn & slip indicator) falls into the turn ... in an uncoordinated turn (slipping
  turn). What are the forces that goes into this?

The ball is displaced to one side because the aircraft is being allowed to fly sideways through the air.  The nose is not aligned with the instantaneous direction of the flight path, but rather is yawed to point to the left or right of the instantaneous direction of the flight path.
As a result, the airflow is striking the side of the aircraft and generating an aerodynamic sideforce, which is what causes the ball to displace to the side.
There are many reasons why an aircraft might tend to fly yawed slightly toward the left or the right in a turn.  Most of these reasons tend to promote a slip, rather than a skid.  Other than the most obvious reason of a significant, intentional deflection of the rudder by the pilot, these reasons include the curving nature of the relative wind during turning flight and its effects on tail surfaces (also described as "aerodynamic damping"), and the fact that the outboard wingtip must move faster, and thus tends to generate more drag, than the inboard wingtip.  In an aircraft with a propeller, P-factor can also play a role, tending to promote either a slip or a skid depending on the direction of turn.  The setting of the rudder trim may also be factor.  When the bank angle is changing, adverse yaw from the deflected ailerons and from the rolling motion itself plays a major role, strongly tending to promote a slip when the bank angle is increasing and a skid when the bank angle is decreasing.  In all cases, the cure is for the pilot to use the rudder as needed to compensate for these effects and align the nose of the aircraft with the instantanous direction of the flight path, so the nose is pointing straight into the "relative wind" and the airflow is not hitting the side of the fuselage.
Note that "having too little or too much lift in relation to the bank angle" or "having too little or too much centrifugal force in relation to the bank angle" are not causes of slips or skids.  Rather, they are results of slips or skids.  The causes are the aerodynamic effects listed above, creating a yaw torque that makes the aircraft fly slightly sideways through the air, despite the generally stabilizing effect of the vertical fin.  More on this later.
So why does the slip-skid ball get displaced off center when the aircraft flies sideways through the air?  When the airflow hits the side of the fuselage, this generates an aerodynamic sideforce, acting orthogonal (perpendicular) to the flight path in the direction that is roughly parallel to the wingspan,  This sideforce points toward the high wingtip in a slip, and toward the low wingtip in a skid.  This real aerodynamic force has been completely omitted from the diagram of the slipping turn attached to your question.  This is a major flaw in the diagram.
This aerodynamic sideforce contributes to the net aerodynamic force generated by the aircraft, so that the net aerodynamic force vector is no longer pointing
straight up" in the aircraft's reference frame, i.e. parallel to the vertical fin.
The slip-skid ball always tends to position itself in the curved glass tube at the point in the tube that is exactly perpendicular to the net aerodynamic force vector.  If the ball is positioned at any other point in the tube, it will "feel" an apparent force to the left or right, and will shift position.  If the net aerodynamic force is not pointing "straight up" in the aircraft's reference frame, then the ball will not position itself at the "bottom", i.e. at the center, of the curved glass tube.
Another way to look at the situation is to consider the apparent "load" acting on the ball, due to the combined effect of weight and "centrifugal force".  The ball always tends to position itself in the curved glass tube at the point in the tube that is exactly perpendicular to the "load" vector.  If the "load" vector is not pointing "straight down" in the aircraft's reference frame, then the ball will not position itself at the "bottom", i.e. at the center, of the curved glass tube.
The key thing to understand is that the apparent "load" felt by the aircraft structure and contents, including the pilot and the slip-skid ball, is nothing more than the mirror image of the real aerodynamic force vector generated by the aircraft.  This is explored in more detail in a several answers to the related ASE question "Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?".
The diagram is misleading because it suggests that some mysterious effect, presumably somehow related to turn rate, yet seemingly unrelated to any real aerodynamic force, is affecting the amount of "centrifugal force" generated by the aircraft or acting on the aircraft.  This is false.  The difference in the magnitude of the "centrifugal force" vector between a "coordinated" turn and a "slipping" turn is due to the way that a sideslip generates a real aerodynamic sideforce vector, as described above.
While it is in fact true that the "load" acting on the ball can be said to be equal to the vector sum of weight and "centrifugal force", this is not a concept that has a lot of explanatory power, unless we understand that the "centrifugal force" is intimately related to the real aerodynamic forces generated by the aircraft.  The "load" vector, which can be said to be the vector sum of weight and "centrifugal force", is nothing more than the mirror image of the real aerodynamic force generated by the aircraft.  The "load" vector is always equal and opposite to the vector representing the real aerodynamic force generated by the aircraft.1
An extreme case of a slipping turn is a forward slip or sideslip (these are actually exactly the same thing, just aimed differently in relation to the target), where the aerodynamic sideforce from the slip is sufficient to bring the turn rate all the way to zero, creating a linear flight path.  Whenever the flight path is linear, the slip-skid ball doubles as bank angle gauge.  The most extreme case of a non-turning (linear) sideslip is vertically-banked sustained knife-edge flight, like we often see at an airshow.  In this case the aerodynamic sideforce vector from the airflow striking the side of the fuselage (plus an additional component due to the engine thrust line being aimed above the horizon) is supporting the entire weight of the aircraft, and the wing's lift vector is zero.  In this case, the "load" vector is simply equal and opposite the aerodynamic sideforce vector from the slip (including the component due the engine thrust), because aerodynamic sideforce vector and the net aerodynamic force vector are one and the same.  The net aerodynamic force vector is equal to the weight of the aircraft but acting in the upward direction, while the "load" vector is equal to weight of the aircraft, acting in the downward direction.
While established in a steady-state turn at a constant turn rate and airspeed, trimmed for hands-off flight, if we then pull aft on the stick to "load up" the wing with "extra" lift to boost the G-load to a higher-than-normal value for the bank angle, or if we push the stick forward to "unload" the wing to decrease the lift vector and drop the G-load to a lower-than-normal value for the bank angle, the turn rate will immediately change, and so will the horizontal component of the "centrifugal force" vector.  The diagram above might tend to lead us to think that the ball will immediately shift off-center.  This is not the case.
In a situation like this, the flight path will curve up or down, introducing a vertical component of "centrifugal force".  The vector sum of weight and centrifugal force will still be "straight down" in the aircraft's reference frame, and the ball will stay centered.
Or to look at it from a much simpler perspective-- when we pull the stick aft or push the stick forward to change the wing's angle-of-attack, we are increasing or decreasing the magnitude of the lift vector, but we aren't introducing any aerodynamic sideforces-- we aren't yawing the fuselage to fly sideways through the air.  Because the net aerodynamic force vector still acts "straight up" in the aircraft's own reference frame-- parallel to the direction that the vertical fin is pointing-- the ball still stays centered.
(For simplicity, we're ignoring the forward or aft components in the net aerodynamic force vector-- the components that wouldn't show up in a head-on view of the aircraft.  The basic explanation here remains the same regardless of whether thrust is exactly in balance with drag, or not.)
Of course, "loading" or "unloading" the wing in this manner during a turn will also drive a change in airspeed, as the flight path curves up or down.  For an aircraft of a given weight at a given bank angle, there is only one value of the lift vector (G-loading) that will yield a steady-state turn at constant airspeed.2

From reading up online, many sources quote that in a slipping turn, the aircraft is banked too much for the rate of turn.

From the content above, you'll see that this isn't always true.  It can be true given certain constraints, but it is not a concept with a lot of explanatory power.
As noted above, if we are established in a stabilized turn, and then we pull the stick back, or push the stick forward, to boost or decrease the lift vector and G-loading, the turn rate immediately changes, but the ball does not shift off-center.3
The idea that (for a turn at a given airspeed) "in a slipping turn, the aircraft is banked too much for the rate of turn" is only true given the constraint that there is zero upward or downward (skyward or earthward) curvature in the flight path.  And the only way4 that (for a given airspeed) we can have a mismatch between the bank angle and turn rate, while not allowing the flight path to curve up and down, is that we are applying the rudder to expose the side of the fuselage to the airflow and generate an aerodynamic sideforce, or we are failing to apply the rudder as needed to cancel out other aerodynamic torques that are tending to expose the side of the fuselage to the airflow.  If we do this, while adding or subtracting power as needed, we can indeed create the following situations.  Starting from a stabilized coordinated turn at a constant airspeed and altitude, we can--
a) vary the turn rate and turn radius while holding altitude, airspeed, and bank angle constant
b) vary the bank angle and while holding altitude, airspeed, and turn rate and radius constant
The ball will deflect off center in both of these sitations.  It is equally true to say that the ball goes off-center because the centrifugal force vector is no longer correctly matched to the bank angle, or because the load vector (the vector sum of centrifugal force and weight) no longer points straight "down" in the aircraft's reference frame (i.e. parallel to the vertical fin), or because the net aerodynamic force vector no longer points straight "up" in the aircraft's reference frame (parallel to the vertical fin), or because the pilot is using the rudder in a way that exposes the side of the fuselage to the airflow, which generates an aerodynamic sideforce.
The pilot's rudder usage is the key to centering the slip-skid ball, or to intentionally driving it off center.  The sideways force we "feel" in a slip is the real aerodynamic force generated by the air striking the side of the fuselage, which changes the direction of the net aerodynamic force vector, and its mirror image, the "load" vector.  Explanations centered around some sort of "balance" or "imbalance" between bank angle and turn rate, or bank angle and lift force, or bank angle and G-loading, are misleading, and have little explanatory power, and certainly do not apply to aerobatic flight.  These explanations tend to obscure, rather than illuminate, what is really going on.
Some of these sorts of explanations are well suited to a car driving on banked track, or a bobsled going down a banked track, but not to flight, where the trajectory is not constrained to have a specific turn radius, and also is not constrained to have no upward or downward (skyward or earthward) curvature.5
Note that in flight, generally speaking, changing airspeed while holding the bank angle constant creates no tendency toward a slip or skid, though the turn rate and radius will both vary.  As long as the net aerodynamic force vector continues to point "straight up" in the aircraft's reference frame, there will be no slip or skid.  Though again, we can come up with specific contrived situations where varying the airspeed while holding the bank angle constant does cause a slip or skid-- for example if the turn rate and radius are constrained to remain constant as the airspeed is changed.  Again, this can only happen if the pilot applies the rudder to expose the side of the fuselage to the airflow, generating an aerodynamic sideforce.

I have trouble understanding why

This is not surprising, because you have been provided with faulty explanations, as well as faulty diagrams.  The diagrams attached to your question have omitted the aerodynamic sideforce vector caused by the airflow striking the side of the fuselage.  The diagrams attached to your question give the impression that the net aerodynamic force vector generated by the aircraft is the same in all 3 cases (coordinated flight, slipping flight, and skidding flight), when it actually is not.  The diagrams attached to your question give the impression that the "load" vector can somehow be something other than the mirror image of the net aerodynamic force vector, when it actually cannot be anything other than the mirror image of the net aerodynamic force vector.
To read more about the specific errors in the vector diagrams attached to your question, see the related question What is missing from these diagrams of the forces in slips and skids? and answer What is missing from these diagrams of the forces in slips and skids? .
Other related ASE questions or answers:
(Q) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?
(A) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?
(Q) What does the balance ball actually indicate?
(A) What does the balance ball actually indicate?
The topic of accuracy of the diagrams illustrated in your question has also been discussed on Physics Stack Exchange.  See--
(Q) Is this vector diagram of the forces at play in turning flight correct?
(A) Is this vector diagram of the forces at play in turning flight correct?
Footnotes
1 -- For simplicity, we're assuming that the slip-skid ball is located near the aircraft's CG, or aircraft's yaw rotational rate is constant, or both.  A change in the yaw rotational rate will create an apparent centrifugal or inertial force that displaces the slip-skid ball to one side if it is located far forward or aft of the CG, and this displacement does not reflect any real aerodynamic force component.  For example, if the yaw rotational rate is increasing toward the left, a slip-skid ball located far in front of the CG would tend to be displaced toward the right, and a slip-skid ball located far aft of the CG would tend to be displaced toward the left.  These second-order effects are beyond the intended scope of this answer, and are also beyond the scope of what pilots are generally able to notice in actual flight.
2 -- For simplicity, we're overlooking the fact that magnitude of the lift vector associated with a steady-state coordinated turn at some particular bank angle and some particular airspeed and angle-of-attack is very slightly different when we are descending or climbing, than when we are maintaining altitude (with respect to the airmass in all cases).  For modest dives and climbs, this is a very minor effect and definitely not something that someone still learning about the basics of slips and skids should concern themselves with, but in the interest of accuracy, this point should not be completely omitted.  To learn more about the "unloading" of the lift vector in a descent or climb, see What produces thrust along the line of flight in a glider? and Does lift equal weight in a climb?
3 -- Surprisingly, Wolfgang Langewiesche's classic flight-mechanics-for-pilots book "Stick and Rudder" (1944), which contains a lot of good material on the physics of turning flight, does make the claim (see pages 205, 219-220, and 223-226) that while excess back pressure on the stick does not cause a slip or skid, too little back pressure on the stick can cause a slip, with the ball deflected to the low side of the turn, which should be corrected by increasing back pressure rather than by applying inside rudder.  This is probably the only faulty idea in the entire book.  The 3rd edition of the book "Modern Airmanship" (1966, edited by Neil Van Sickle, Major General USAF) has similar content on pages 308-309.  By the 8th (1999) edition of "Van Sickle’s Modern Airmanship", edited by John F. Welch, Lewis Bjork, and Linda Bjork, this content had been deleted, while the remaining content on the physics of turning flight had been retained (page 441).  For more, including specific quotes from these books, see https://web.archive.org/web/20180905112047/http://aeroexperiments.org/critiques.shtml .
4 -- Strictly speaking, it's not exactly true that the only way that (for a given airspeed) we can have a mismatch between the bank angle and turn rate, while not allowing the flight path to curve up and down, is to deflect the rudder to the side in a way that exposes the side of the fuselage to the airflow.  But it's a good first approximation.  For a more nuanced view, we have to consider the effect of the sideforce generated by the rudder itself.  For example, we could shut down one engine on a conventional piston twin and apply the rudder just enough to keep the fuselage exactly aligned with the airflow.  Since the sideforce from the rudder itself is not zero, the ball will be slightly deflected (toward the working engine) when the nose of the aircraft is pointing straight into the relative wind.  If we bank into the good engine as needed to cancel all turning tendency, we'll end up flying along in a straight line (zero turn rate) while banked slightly toward the good engine, as discussed in detail elsewhere on ASE, such as here, here, and here (link to be added).
5 -- Speaking of bobslebs on curved tracks, "Hang Gliding Training Manual" and "Performance Flying" by Dennis Pagen both include extensive content-- including an illustration of a bobsled on a track-- purporting to show how, as a pilot enters a turn while flying a hang glider, if he or she doesn't immediately "load up" the wing with an adequate pitch input, the glider will slip sideways through the air, toward the low wingtip.  Also the idea that an intentional reduction of angle-of-attack and lift and G-loading while turning will make the glider slip sideways through the air toward the low wingtip.  See for example pp. 128-129 in "Hang Gliding Training Manual", and page 45 in "Performance Flying".  A similar idea appears on page 70 of Peter Cheney's "Hang Gliding for Beginner Pilots" (3rd edition, 1997).  I've checked these ideas in flight with a yaw string, in hang gliders as well as conventional aircraft, and found no basis for them.  For more on the content in these books, see https://web.archive.org/web/20180905112047/http://aeroexperiments.org/critiques.shtml .

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Because of the insufficient speed and pull for the bank angle of the aircraft  .For level  turn  you must pull up not bank  ,but first you need to bank some degrees and have enough speed to maintain the G required for the bank angle .If you pull lower than the needed G's for bank angle your aircraft nose will dive ,if you pull more G's the nose rise above horizontal line.Now there are two types of skidding turns (inside the turn or  drifting outside of  turn)

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What causes the ball to not be centered in a slip/skid?
I've reviewed quite a few questions on this forum including
Forces in a slipping turn
is-this-vector-diagram-of-the-forces-at-play-in-turning-flight-correct
what-does-the-balance-ball-actually-indicate
but I don't believe they have answered this specific question.

Btw, if this pertains to helicopters, the sideforce from the tail rotor is significant and can cause the slip-skid ball to be off center even when the yaw string is perfectly centered (fuselage streamlined to airflow). Pertains to the paragraph in my answer starting with "All the above is really just a first approximation, but it is a very good approximation for most situations in actual flight." So it appears that in a helicopter in cruising flight, the most efficient situation is to have the yaw string centered but slip-skid ball slightly off center,  meaning that the a/c is slightly banked.

Of course it goes without saying that any time the flight path is linear, the slip-skid ball doubles as a bank angle gauge.

PS I'm not a helicopter pilot, just thinking through the physics--

It is the same thing that makes you slide sideways in your seat going around a corner in your car.
Or the same thing that would cause you to slide sideways the other way if you were driving straight ahead on a side hill.
Don't overthink it... the ball simply responds to the same forces you feel.
ADDENDUM:
“I agree the comparisons are spot on if you are only considering a coordinated turn, but I'm asking about a steady level turn in which the aircraft is slipping/skidding; which I don't think the comparison can easily be extended to?”
I actually think it is the other way around.  Consider the following:
•   CAR - Unless it is travelling on a banked racetrack, at precisely the speed the banked turn was designed for, a car will never be able to make a “coordinated turn”.  There will generally always be a centrifugal sideforce.
•   PLANE - Once adverse yaw at the turn initiation has been countered, and when established in a constant angle of bank turn, most (positively stable) aircraft will default to a coordinated turn.
In case it helps further, let’s take the aircraft “uncoordinated turn” to a couple logical extremes:

Wings Level, unbanked turn, using rudder only.  Think about this as it compares to turning a car on level ground.  This is an exaggerated skidding turn, with all the centrifugal force pulling you (and the ball!) horizontally, or sideways.

Straight ahead, level flight in a side slip.  The aircraft nose is tracking straight ahead and not turning, while the wings are banked.  This is an exaggerated slipping turn, (minus actually turning...) and is just like the example of driving your car on a side hill.  In this case, gravity is pulling you (and the ball!) vertically straight down, which from the aircraft's frame of reference, is a bit sideways.

The ball is a "dumb" instrument.  Nothing more than ball in a tube reacting to centrifugal force and gravity.  You could accomplish the same thing by holding a full glass of water between your knees:  If the inside knee gets wet it is a slipping turn, if the outside knee gets wet it is a skidding turn.
If these explanations are inadequate the only thing left to suggest is that you find a local flight school that offers orientation flights and ask the pilot to demonstrate for you.
If that suggestion isn't practical or affordable, try this simple experiment:  Get a round mixing bowl and a marble.  The round bowl simulates the curve of the glass tube.  Put the marble in the bowl and hold it at arms length and begin turning in place.  (if you think you might get dizzy and fall against the furniture, wear a helmet!)  Your goal is to adjust the angle of the bowl to keep the marble in the center.  Note that if you tip the bowl too much either way the marble will start to roll up the sides.
It isn't a perfect comparison because it doesn't simulate yaw, but can you appreciate the correlation?  Please let me know if any further explanation is needed...

Thanks for the reply @Michael Hall! But I don't see how this answers my question as the dynamic of a car (ground friction) are different than the dynamics of an airplane (aerodynamics)?

True that some "dynamics" are different, but other fundamental forces are identical.  This is one of them, and the comparison is entirely valid.  Centriifugal force and gravity, that's it.  As I cautioned, don't overthink this just because it is an airplane, and because other explanations include graphs and diagrams...

@eball Replace ground friction with the side force from a tilted wing and the comparisons are spot on. +1

@MichaelHall, I agree the comparisons are spot on if you are only considering a coordinated turn, but I'm asking about a steady level turn in which the aircraft is slipping/skidding; which I don't think the comparison can easily be extended to?

@eball, please see my addendum.

@MichaelHall, thanks for the addendum! It definitely helps clear some things up. But I wanna summarize what you said. Focusing on your first PLANE example when you are turning to the left, you need to add left rudder which will put a force to the right on the tail causing you to yaw to the left putting you in a skid. If you had water between your legs, your right leg would get wet. The only difference between this and a car is that the car cannot have a sideslip angle?

@eball, I'm not totally clear on what you are asking, but the goal of the pilot is to coordinate just enough rudder rolling into turn initiation so that the ball never moves, and water never sloshes.  To prevent a slip or skid.  And I believe that in a turn a car does have a sideslip angle.  It is countered mostly by the friction of the tires.

@eball I think any time the bank angle of the track is not perfectly correct for the car's speed, the ground is pushing in some direction other than "straight up" in the car's reference frame, meaning that lateral friction is playing a role.  If we looked at the physics here in detail I think we'd find that there is a very small, but non-zero, sideslip angle, between the direction the tires are pointing and the direction the car is actually moving.  I may be wrong about that though-- maybe it's just static friction between the tire and the ground.

@MichaelHall, I've thought about it a lot more and I think that you're right. If you consider the angle of the banked turn to be analogous to the bank angle of the helicopter, a parallel can be drawn between the slip angle of the tire and the sideslip angle of the helicopter. Previously I wasn't considering the tires as something that could slip; I was only considering the fuselage.

@quietflyer, I think you are right! But I don't believe the slip angle of the tire needs to be small (at least that's my current understanding of how tires work/are modeled).

@eball -- maybe what is happening is that the contact patch of the tire is not slipping/skidding relative to the ground, but the tire is actually twisted and that's how a lateral force can be generated--

Assume the aircraft is trying to perform a steady, level, coordinated turn. If the airplane is slipping, excess outward lateral force (${\bf L}_{slip}$) will be generated. To be coordinated, the bank angle must be decreased or the rudder into the turn increased.
Because the ball will position itself such that it is normal to the sum of the external forces in a coordinated turn (here simply the lift vector $\bf L$, it should sit in the center, in a coordinated turn (similar to a banked turn in a car where the normal force applied by the road replaces the lift force).
In a slip, the ball will be to the left of center because it rests at the point where the surface is normal to the Lift vector required for a coordinated turn at the same turn radius and airspeed (equivalent to the sum of the lift vector $\bf L$ and ${\bf L}_{slip}$) as indicated in the diagram.
The old adage "step on the ball" applies here.
The below diagrams should help clarify:

Thanks for the edits @Robert DiGiovanni!! I saw you were an active contributor in the other threads I linked so I value your input. But why would the airplane generate insufficient lateral outward force? Looking at the diagram, wouldn't it generate excess outward lateral force because the fuselage acts like a wing at an angle of attack of $\beta$ (which would need to be balanced by a larger inward force in a turn, provided by a larger bank angle $\phi$)? -Thanks!

the "ball" inclinometer is in the aircraft reference.  Wing does pull the plane sideways, but needs help to center the ball.  Thats what the rudder does.  Remember they are teaching shallow turns, and adverse yaw of the ailerons also needs to be counteracted.  "Step on the Ball".  Adding more bank just makes it harder to center, as the center is now more "uphill".

Ooo, I think we may be talking about this from the opposite point of view. I'm trying to figure out how to fly with sideslip (and it's effect on the ball), not how to recover. So If you are in a steady slip, you need to decrease the angle of bank to return to a coordinated turn. But if you are in a coordinated turn and want to slip, you need to increase the angle of bank. Right? (I believe they way my answer above is currently written disagrees with the diagrams below it haha.) Thanks!

yes, increase the bank and/or push the rudder the other way.  That's how you do a side slip (against a cross wind), but please make sure you have adequate airspeed and expert instruction as flying uncoordinated can have a bad ending.  What you do with the controls depends on what you want the plane to do.

Thanks for clarifying @RobertDiGiovanni! I've edited the answer to capture what we've discussed. And thank you for the concern, but I'm not a pilot haha. I have limited stick time in simulators, but am trying to understand this to help develop a pilot model. Thanks again for the help!

A suggestion-- in the diagrams, the apparent inertial force "centrifugal force" (which is a pseudoforce) is equal and opposite the net centripetal force (turning force), which is equal to the vector sum of the (leftwards) horizontal component of the lift vector plus the (rightwards) horizontal component of Lslip.

In the slipping diagram, it looks like you need to make the Lslip force much larger (and pointing to the right and slightly upward, perpendicular to the wing's lift force) to balance the strong leftward force component due to the steep bank angle, to end up with the same apparent "centrifugal force" vector as in the coordinated turn at the same airspeed and turn radius.

Also, one of your sentences got me thinking-- and ended up motivating the last paragraph now added to my answer.  I think the sentence could be improved by adding the words in italics: "In a slip, the ball will be to the left of center because it rests at the point where the surface is normal to the Lift vector required for a coordinated turn at that same turn radius and airspeed (equivalent to the sum of the lift vector L and Lslip) as indicated in the diagram."

Without this addition, readers may tend to imagine a comparison with a coordinated turn at the existing bank angle, which isn't valid.  Or if you prefer, the addition could be at that same turn rate and airspeed.

@quietflyer, my diagrams could definitely use some TLC haha, but I do believe that the Lslip vector should be drawn perpendicular to the velocity vector (assuming ideal flow, i.e. incompressible and inviscid).
I think that edit does help clarify the situation so I'll add it, thanks for the suggestion!

A simple answer: the ball is forced to move along an arc in a plane. Let's imagine the center of that arc as the point where the resultant of the sum of the gravitational and inertial forces is applied. Now, you project that resultant on the plane of that arc. The intersection of the arc and the projected resultant is exactly where the ball lies, at any time or circumstance...

See my second-to-last paragraph for similar ideas (but not explained so concisely), which itself was inspired by something in eball's answer, namely the concept that the ball always sits at the point in tube that is "normal" to the lift vector that would be required in a coordinated turn at that same velocity and turn rate or turn radius--

(now third-to-last paragraph)

I hope this eventually becomes the "accepted" answer, very nicely put.

Wow, @xxavier, that is a really concise/correct way to explain it! You could also say "Let's imagine the center of that arc as the point where the resultant of the sum of the external forces ($\bf L$ and ${\bf L}_{slip}$) is applied."

@quietflyer, I would make this the accepted answer but it doesn't address where ${\bf L}_{slip}$ comes from...tho I guess that's irrelevant if you think of it from the point of view xxavier did above (i.e. the gravitational/inertial forces instead of the external forces)

As the OP is not a pilot (but asking a very good question), it remains to be said that flight is 3 dimensional and adequate vertical lift is always required to do a ... "steady level turn".
So now we may discuss the virtues of flying the plane, not the instrument.
It so happens that increasing bank, or slipping, is an undesirable turning technique because it requires more pitch (or more speed) to maintain adequate vertical lift.  You now have a double whammy of airspeed reducing drag factors.  What happens to a pilot at night or in the clouds?  They start pulling back on the stick more, losing airspeed and rapidly entering a spiral.
Skidding flight also creates a higher drag condition, and also carries the risk of tip stalling the slower wing.
This is why "low and slow" planes benefit by staying coordinated, but exactly what is this?
For cars, trucks, and planes alike it is the direction of inertial resistance to the vector sum of gravitational and aerodynamic acceleration directly towards the bottom of the vehicle.
For a turn of a given radius and bank, the controlling factor is speed.  This works for cars and certainly for trucks.
With aircraft, bank is rarely given, but radius and speed can be (when flying a landing pattern curcuit).  When speed is given, we match bank (ailerons), and "steering" (rudder), to hold course in a coordinated (lowest drag) manner while also providing adequate lift for level flight.
So, pilots do not choose (even though it is theoretically possible) to control coordinated flight by changing bank.  Far better to reduce drag with rudder by  stepping on the ball.  As drag will always be higher in turns than straight flight, increasing power to maintain airspeed is also not a bad idea, especially if your plane is very draggy.
So it follows that adequate airspeed is crucial for safe flight and lower drag helps maintain airspeed, so slipping (or skidding) in level turns is out (not safe).
These techniques do have value in adjusting glide angles for landing but should not be attempted without understanding the risks.

You're a little late to the party... Good answer, but now that we know he's writing software for a helicopter simulation I'm not sure this is helpful.

@Michael Hall  well, I felt a little responsible for all this, but hey, would this not apply to overbanking a helicopter too?

Helicopters are a whole different animal.  I only flew one once, but tail rotors and power changes magnify what us fixed wing pilots experience WRT the ball by 10x...

@RobertDiGiovanni, why is it more efficient to step on the ball than to change bank? Is this because to change bank at a given speed/turn radius, you have to increase AOA which is more draggy than using rudder to gain the inward force required for the turn?

@eball the actual difference will vary from aircraft to aircraft, but the answer is yes.  With increase bank one must increase pitch to maintain vertical lift...more drag.  Ball is not centered. Also more drag.  The gist of your idea makes sense, as the horizontal lift vector of a banked wing will produce a lot of lateral force.  But here, it is acceleration  that holds the ball "up", and the drag of a wing moving sideways will not allow you to accelerate for long in one direction.  The rudder keeps changing that direction.  Acceleration from 0 (constantly) is easiest.

The ball is off-center in a slip or skid because the bank angle is "wrong" for the turn rate and airspeed, and thus the resultant of "centrifugal force" plus weight does not point "straight down" in the aircraft's reference frame.
While true, this statement isn't terribly useful from a pilot's point of view, since we normally don't vary the bank angle without also changing the turn rate, so we normally wouldn't try to center the ball by adjusting the bank angle.  (An exception would be if we're performing a standard-rate turn by reference to a turn rate indicator or turn coordinator, especially if we're using the old "needle-ball-airspeed" technique, where pilots were advised to keep the turn rate constant with the rudder, while using the ailerons to center the slip-skid ball.)1
An exactly equivalent-- but more useful-- statement is that the ball is off-center whenever the net aerodynamic force generated by the aircraft has a sideways component in the aircraft's reference frame, rather than acting "straight up" in the aircraft's reference frame.
Note that the "centrifugal" force we're speaking of here is actually a pseudoforce.  It is the apparent inertial force generated by the turn.  It is exactly equal in magnitude and opposite in direction to the actual net centripetal force that is causing the turn in the first place.  In a steady-state coordinated turn, this centripetal force is simply the horizontal component of the lift from the banked wing, while in an uncoordinated turn, we have to add or subtract an additional sideways force that is caused by the fuselage moving sideways through the air.
The analogy of a car going around a banked track can help us understand all this.  In this case, substitute the force exerted by the track on the tires for the net aerodynamic force generated by the aircraft.  At some combination of turn rate and speed, the banked ground will be pushing "straight up" on the tires in the car's reference frame, and a slip-skid ball would be centered.  In such a case, the tires are "feeling" no sideways loading.  This is like an airplane in a coordinated turn, where the net aerodynamic force acting on the aircraft is simply the lift force from the wing, acting "straight up" in the aircraft's own reference frame.
However, it's important to recognize that the car analogy only goes so far.  After all, the track is constrained to have a certain turn radius regardless of the car's speed, and that's not true in actual flight.
It's easy to see why at all speeds except for one, the bank angle will be "wrong" for the turn rate as a car goes around a banked track.  But why would the bank angle ever be "wrong" for the turn rate and airspeed in actual flight?  Or to put it another way, why would the net aerodynamic force acting on the aircraft ever be anything other than the lift force from the wing, acting "straight up" in the aircraft's own reference frame?
Beware of explanations that invoke "centrifugal force" without giving sufficient attention to the real, tangible, aerodynamic forces generated by the aircraft.
At least one aviation author (writing for pilots of ultralight aircraft) has actually suggested that flying at the "wrong airspeed for the bank angle" will make the aircraft slip or skid.  This is completely wrong, unless we artificially constrain the turn radius to be fixed, which is a very odd constraint in the context of flight.
So in actual flight, how can it happen that the bank angle can be "wrong" for the turn rate and airspeed?
Because the aircraft is being allowed to fly sideways through the air-- i.e. the nose is not being kept pointed in the same direction as the aircraft is actually travelling through the air at any given instant.
As a result, the airflow is striking the side of the fuselage and making a real, tangible, aerodynamic sideforce, acting to the "left" or the "right" in the reference frame of the aircraft and pilot.  The slip-skid ball "feels" the combined effect of this sideforce, and the wing's lift force.
Or to put it another way, this real, tangible aerodynamic sideforce adds or subtracts to the turn rate, so that the turn rate is no longer "right" for the bank angle and airspeed, so the slip-skid ball goes off-center.
The ball will always displace in the "upwind" direction relative to the airflow that the fuselage is feeling.  It will move toward the side of the fuselage that the airflow is striking.
(An extreme case is sustained, vertically-banked, linear "knife edge" flight, where the wing is unloaded to the zero-lift condition, and the airflow striking the side of the fuselage must support the entire weight of the aircraft, apart from whatever component is supported by the thrust vector.  The ball will be fully deflected, to the "earthward" corner of the curved glass tube.)
So it's equally valid to say that the slip-skid ball will be displaced off-center whenever the bank angle is "wrong" for the turn rate and airspeed, or to say that the slip-skid ball will be displaced off-center whenever the airflow is striking the side of the fuselage because the nose of the aircraft is not aligned with the actual direction of travel and the airflow is hitting the side of the fuselage.  The two statements are equivalent--they are two different "sides of the same coin".  Pilots who understand that, understand what the ball is really telling them.
Once we understand this, it becomes clear why in aircraft with no propwash over the nose-- such as sailplanes-- a yaw string (a tuft of yarn attached to the outside of the canopy, on the aircraft centerline, in the pilot's field of forward vision) is a perfectly acceptable substitute for a slip-skid ball.  Any discrepancy between the direction the aircraft is pointing and the direction the aircraft is moving will make the "yaw string" blow sideways.
Whenever the pilot uses the rudder as needed to keep the nose of the aircraft pointing directly into the airflow, so that there is no airflow striking the side of the aircraft, the ball (and the pilot) only "feel" the lift generated by the wing.  Since the wing's lift force always acts "straight up" in the aircraft's reference frame, the ball will always be centered in such a case, regardless of the bank angle angle or airspeed.  In such a case, the turn rate will automatically be "correct" for the bank angle and airspeed.
Note that if we change the shape of the fuselage, we also change the amount of deflection of the slip-skid ball that we'll see for some given sideways angle of airflow.  A slender streamlined fuselage will create less sideways aerodynamic force, and less displacement of the slip-skid ball, than a big boxy one.
The biggest deficiency in diagrams like the ones accompanying this related ASE question is that they don't specifically show the aerodynamic force created by the airflow striking the side of the fuselage in a slip or skid, and thus leave the reader with no understanding of what is really going on in a slip or a skid.
All the above is really just a first approximation, but it is a very good approximation for most situations in actual flight.  To take the conversation to a deeper level, we can recognize that it's actually possible for the rudder itself to create a significant sideforce--which will affect the turn rate, and will also affect the slip-skid ball--even if the fuselage is completely streamlined to the airflow.  That's why the most efficient flight is not always obtained with the slip-skid ball completely centered.  This comes into prominence when we are dealing with a failed engine on a twin-engine aircraft, because the rudder will be strongly deflected.  But for all practical purposes, as long as we are not dealing with a huge asymmetry in thrust, it's a very good approximation to  assume that whenever the ball is off-center, it's because the pilot is not keeping the nose pointed in the same direction as the aircraft is actually travelling, and thus is allowing the airflow to strike the side of the aircraft.
Please keep in mind that the "airflow" we're speaking of here has nothing to do with the external, meteorological wind.  Rather, it is the "relative wind" created by the aircraft's own motion through the airmass.  When the aircraft is not pointing in exactly the same direction as it is actually travelling through the airmass, this "relative wind" has a sideways component in relation to the fuselage.  The pilot fixes this by applying the rudder as needed to keep the aircraft pointing in the same direction as it is actually travelling.  The rule is to "step on the ball" to bring it back to center.
We could go on to discuss the reasons why many aircraft-- especially slow-flying, long-spanned aircraft like sailplanes-- tend to require the pilot to hold a touch of inside rudder during a turn to keep the slip-skid ball centered.  But that is better saved for another question.  For now, we'll simply note that while the vertical fin does tend to keep the nose of the aircraft pointing directly into the airflow, like a weathervane, sometimes it needs a little extra help from the rudder.
What if we are maneuvering in three dimensions rather than just two?  The concept of the slip-skid ball going off-center whenever the bank angle is "wrong" for the turn rate and airspeed is really only valid when there is no vertical acceleration.  For example, near the top of a wingover, the bank angle may briefly approach vertical and the airspeed may drop quite low, yet given appropriate rudder usage, the ball can remain centered.  A more complete statement would be that the slip-skid ball will go off-center when there is a mismatch between the bank angle, turn rate, airspeed, and vertical acceleration component.  The slip-skid ball will position itself in the tube to be aligned with the vector sum of weight, the horizontal "centrifugal force" pseudoforce generated by the turn, and a vertical (upward) "centrifugal force" pseudoforce generated by the earthward curvature in the flight path.  But even in this more complex case, it's still true to say that if the aircraft is kept aligned with the instantaneous direction of travel so the airflow is not allowed to strike the side of the fuselage, the slip-skid ball will stay centered.
Here's an interesting point originally brought out in another answer (and now also in yet another)-- we can imagine that the ball represents an imaginary pendulum, suspended from a point located at the center of the radius of curvature of the glass tube.  For turns where we are only maneuvering in two dimensions (as opposed to the wingover example above), if we varied the bank angle while holding the airspeed and turn rate constant, we'd notice that the position of the pendulum (or the position of the ball) relative to the horizon would stay the same.  The ball always moves to the part of the tube that is "normal" to the vector sum of the weight vector and the apparent "centrifugal force" (pseudoforce) vector.  This also means that the ball always moves to the part of the tube that is "normal" to the vector sum of upward lift and net centripetal force.  And this also means that the ball always moves to the part of the tube that is "normal" to the direction of the lift vector that would be required for a perfectly coordinated turn at the current airspeed and turn rate.  So we can think of the ball as telling us what bank angle would be required for a perfectly coordinated turn at the current airspeed and turn rate.  This brings us back around to the opening line of this answer-- the concept that in a slip or skid, the bank angle is "wrong" for the turn rate and airspeed.  In theory, in a slip, we can center the ball by reducing the bank angle while holding the turn rate and airspeed constant.
But unless we're performing a standard-rate turn by reference to a turn rate indicator or turn coordinator, this isn't a terribly useful concept in actual flight, since any change in bank angle will normally also drive a change in turn rate.  Rather than thinking of adjusting the bank angle to match the airspeed and turn rate, thus centering the ball, it's much more useful to recognize that the bank angle will always be "correct" for the airspeed and turn rate as long as the airplane is not being flown sideways through the air. That's why, practically speaking, the rudder is almost always the best tool for centering the slip-skid ball.
In fact, in the few instances where we choose to use the ailerons rather than the rudder to center the ball, the main reason that works is not because we are changing the bank angle, but rather because the ailerons are generating an "adverse yaw" torque that has the effect of a rudder input in the opposite direction.  Example: we're holding a thermal turn in a sailplane and we're suspecting that a steeper bank angle might help us to stay in the thermal better-- the ball accidentally moves to outside (high side) showing a skid-- we know that applying outside (high side) rudder would center the ball, but also will reduce the bank angle due to coupling between yaw, slip, and roll-- but we don't want to reduce the bank angle as we might lose the thermal-- so we apply inside aileron instead.  The main reason that this centers the ball is actually not that the bank angle is increasing, but rather that the ailerons are creating an "adverse yaw" torque and we aren't doing anything with the rudder to counteract it.  So the nose swings away from the direction that we've moved the aileron control stick, bringing the nose into better alignment with our actual direction of travel, and increasing the bank angle as an extra bonus. In actual practice, in an aircraft with a high wing or dihedral or sweep, if we want to make a yaw adjustment (i.e. change the sideslip angle) without introducing a roll torque, we actually need to move the rudder one direction while also moving the ailerons slightly in the opposite direction.  There are situations where it will best suit our purposes to use only the rudder ("step on the ball") or only the ailerons ("move the stick against the ball") for a correction, but the former is much more common than the latter, and the latter works mainly due to "adverse yaw" from the ailerons, not as a direct result of the actual change in bank angle.
Footnotes--

As Wolfgange Langewiesche noted back in 1947 in his classic book "Stick and Rudder", this very antiquated "needle-ball-airspeed" technique is arguably rather flawed.  It would fail completely on any aircraft that didn't either a) exhibit substantial coupling between yaw, sideslip, and roll, or b) generate substantial aerodynamic sideforce during a sideslip or skid.  It's better to control to the turn rate with the ailerons, while using the rudder as needed to center the slip-skid ball-- this will work in any aircraft.  Nonetheless, this old technique is likely what led to a turn rate indicator (with included slip-skid ball) sometimes being a called a "turn and bank indicator", as if the indication of the ball somehow had something to do with the bank angle.

thanks for the detailed reply! I'm currently working on helicopter simulations and have never actually flown a real aircraft so the controls are not intuitive to me. But your explanation seems correct to me. And I was actually looking for those diagrams you said you'd get to eventually! haha. Thanks again for your input.

Sometimes it’s hard to know whether you are answering for a teenager who’s never taken a hard corner, unbelted, on a bench seat, or a computer programmer looking for equations...  (doubly confusing if you don’t know what “unbelted” or “bench seat” means!)

Haha totally agree!! Probably would have been helpful for me to make that clear in my question.

@eball as you are working on simulations of a special breed of aircraft, the helicopters, you shoud know that a helicopter pilot that has flown an entire flight (with a helicopter) so that the ball remained fully centered, is superhuman

There's a duplicated word in this sentence, but I'll wait till I have other changes to make before editing-- " Since the wing's lift force always acts "straight up" in the aircraft's reference frame, the ball will always be centered in such a case, regardless of the bank angle angle or airspeed. "

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is the additional planform area added to an A380 on take-off flaps setting (Flaps 1+F)
I would like to know how I can approximate the additional area added to wing on flaps 1+F:

Can someone use other dimensions they know of the A380 wing and approximate what this area would be, we can say that these 3 sections are rectangular and we can also add 20% to estimate the area added by slats which we can't see in the video.
It would be ideal if I knew the chord length and then I could use the logger pro software to find the width of the rectangle but I have been unable to find this data, even from Airbus because the A380 project is closed.
link to video:

Please someone help me, with this information I will able to do all the maths I am doing using related rates to model the rate of change of lift of an A380 after take-off (basically this additional area will help me get dA / dt where A is area and t is time, it's useful because its a related rate to lift as Planform area is in the lift equation.)
I have tried for days and not been able to get this number, I would appreciate any assistance from you online folks. Even any approximate number is useful please, just get me a reasonable number and also explain how you did it.

You're coming at this wrong.  Fortunately, the world is more simple than you're trying to make it.
Although the wing area can be considered to change (when flaps are deployed), the wing reference area is held constant.
$L = C_L\,q\,S_\mathrm{ref}$
So, the effect of flaps is actually absorbed into the definition of the lift coefficient.  If you think of $C_L$ as a function, it is a function of...
$C_L=f(\alpha,M,Re, \mathrm{Config})$
Angle of attack, Mach, Reynolds number, and Configuration -- i.e. flap position.
If you look closely at $C_L$ vs. $\alpha$ data for a fowler flap system, you will see that there is an apparent increase in the lift curve slope.  That isn't really what has happened -- instead, it is really due to the increase in chord, but (for 2D data) the reference chord is left alone.
There is still a zero-lift angle, but for every degree, the airfoil with a longer chord gets more lift.  So this shows up as an increase in lift curve slope.
If you have detailed aerodynamic data for clean vs. flaps (lift curves and drag polars), then you can certainly use that during your takeoff / landing model.
Usually, the flap position does not change during the takeoff or landing ground roll.
For an aircraft performance and sizing perspective, we treat the deployment of flaps as instantaneous and ignore any time it would take.
I suppose if you were developing a detailed flight simulator and you knew an aircraft had a slow flap deployment/retraction, you could interpolate between the data tables as the flaps move -- but I don't think anyone would include a  $\frac{d \delta_\mathrm{flap} }{dt}$ term.


text:
Thrust needed to overcome parasite drag that increases with speed
Can anyone explain this passage from page 11-5 of the PHAK: “ For example, if an aircraft in a steady flight condition at 100 knots is then accelerated to 200 knots, the parasite drag becomes four times as great, but the power required to overcome that drag is eight times the original value.” Would not the increase in power needed to overcome the added parasite drag be proportional to the amount of that added parasite drag? Why is the power needed 8 times the original value rather than 4?

Drag is a force.  Thrust is a force.  The extra thrust needed to overcome the added parasite drag is proportional to the amount of that added parasite drag.  But power is proportional to thrust times airspeed, and we've doubled the airspeed.


text:
Is the DC-6 Supercharged? And what is a Turbosupercharger?
What it says on the tin. Is the DC-6 supercharged? I thought all planes that fly above around 10,000-15,000 feet need forced induction, but wikipedia doesn't list the DC-6s engines as supercharged or Turbocharged. There was a variant for the P-47 which was "Turbosupercharged"(???? Is that like a Turbocharger and Supercharger in sequence, so air from teh turbo is rammed through the Supercharger?), but not the DC-6 variant.
What gives?

I'm pretty sure turbocharger and turbosupercharger are two different words for the same thing.

Re " but wikipedia doesn't list the DC-6s engines as supercharged or Turbocharged." -- just curious-- why do you say that?  DC-6 page -- https://en.wikipedia.org/wiki/Douglas_DC-6 -- the one embedded link that pertains directly to engines directs to here https://en.wikipedia.org/wiki/Pratt_%26_Whitney_R-2800_Double_Wasp -- go to the "Variants" section https://en.wikipedia.org/wiki/Pratt_%26_Whitney_R-2800_Double_Wasp#Variants and there are plenty of references to supercharging --

Re the last part of your question-- it's actually a duplicate of a question I asked myself.  Had forgotten all about that.  https://aviation.stackexchange.com/q/71210/34686

Yes, the DC-6 is supercharged with mechanically driven centrifugal superchargers.  A “turbosupercharger” is another name for “turbocharger” which is an exhaust driven centrifugal supercharger.
The DC-6 uses Pratt & Whitney R-2800 engines. The R-2800 has a mechanically driven centrifugal supercharger at the back of the engine. This is common for most WWII and later radial engines.
Some other WWII aircraft (B-17, B-24, B-29, P-47, etc.) also added an exhaust driven turbocharger (also called turbo supercharger) in addition to the centrifugal supercharger for additional boost and high altitude performance.
Wikipedia photo Pratt & Whitney R-2800 Double Wasp

your answer is the only one that actually was an answer to the question! +1 from niels

From the Wikipedia entry on "turbocharger" (bolding added)1:

The current categorisation is that a turbocharger is powered by the
kinetic energy of the exhaust gases, whereas a supercharger is
mechanically powered (usually by a belt from the engine's crankshaft).
However, up until the mid-20th century, a turbocharger was called a
"turbosupercharger" and was considered a type of supercharger.

In actual fact, on an aviation engine, a supercharger that did not involve an exhaust gas turbine would almost always be gear-driven, not belt-driven.
Also, it appears that the usage of the term "supercharger" to refer to the larger category of all forced induction systems, whether gear-driven or driven by an exhaust gas turbine, has lingered longer in the aviation world than in the automotive world.  Thus you still may encounter the terms "gear-driven supercharger" and "turbosupercharger".  The latter term is synonymous with "turbocharger".
In the aviation context, even when consulting modern sources, you should generally not assume that the word "supercharger" always implies a device that is driven by gearing to the crankshaft rather than by a turbine in the engine exhaust gas flow. It may in fact refer to a turbosupercharger, i.e. a turbocharger.2
Some aircraft with turbosuperchargers (i.e. turbochargers) also have gear-driven superchargers.  This was true of nearly all, or all, of the World-War 2-era combat aircraft with turbosuperchargers, such as all variants of the P-47 Thunderbolt.3  But the usage of the word "turbosupercharger" is not meant to specifically signify that a gear-driven supercharger is also present.4
In the diagram below of a World-War-2-era supercharger installation, the "internal blower" (item 3) is the gear-driven supercharger.

Source of image: this page from the website of the Aircraft Engine Historical Society
Another answer has already addressed the specifics of the (gear-driven) supercharging system used on the DC-6, so I'll end here.
Footnotes:

Accessed 7/28/23

This terminology is discussed in more detail in answers to this related ASE question: In the modern aviation context, is the word "supercharger" generally assumed to mean a gear-driven system rather than an exhaust-driven system?

See the related ASE question Did any WWII aircraft only use a turbocharger but not a supercharger? .  Also, for more insight on the advantages of combining turbosuperchargers (turbochargers) with gear-driven superchargers in high-performance World-War-2 era aircraft engines, see this 1943 document from General Electric: "The Turbosupercharger and the Airplane Power Plant".  And here's a "Quora" post that also gives some insight into the topic, rather concisely.

In other words, the word "turbosupercharger" should not be considered to be a "portmanteau" of "turbocharger" and "supercharger", signifying that both features are present; rather it is a simply the combination of "turbo" and "supercharger", describing the type of supercharging that is present.

Really it's quite logical that way. To "supercharge" means to increase the "charge" (of air and fuel), and needn't inherently say anything about how that's done. A "turbosupercharger" is a supercharger driven by a turbine. "Turbocharger" is just a reduced version of "turbosupercharger" that's less of a mouthfull.

Linguistically, "supercharge" and "overload" are doublets (the former Latin via French, the latter Germanic).

Just to add a little more confusion (and amazing engineering)...  Some piston engines were 'turbo-compounded'.
This is when they used a turbine in the exhaust gas to add shaft power back to the engine crankshaft.  These engines also had a shaft-driven supercharger.
The TC18 variants of the Wright 3350 were the most widely produced and used turbo compound engines ever produced.

By the way, the centrifugal blower at the back of a radial engine served two purposes: it stirred up the fuel and air mixture to produce a uniform blend which was then presented to each intake runner tube in the engine, and it provided pressure boost to the runners as well.
The boost function could be tailored for the specific application by swapping out the drive gears; a higher gear ratio spun the impeller faster and furnished more boost, and better high-altitude performance.
Although not a radial, the Rolls-Royce Merlin engine had a mechanical boost with a two-speed gear drive, selectable by the pilot: low boost for lower altitude use, and high boost for high altitude use. This gave it superior high-altitude performance.


text:
How elevator and ailerons affect attack and roll angles?
I was trying to model the flight of an aircraft and i need some function that expresses the change of attack and roll angle in function of the elevator and ailerons (respectively).
Does anyone have an idea of how is this function or where i can find some data to interpolate?
Thanks to all those ones that will help me!

The answer to this would be extremely variable based on the airfoils used for the wing & horizontal stabilizer. Pitch and roll rates achieved, and required, in an acrobatic aircraft would be highly undesirable in a basic trainer. Without narrowing the scope of the question, the only possible answer is, "it depends."

Although for a basic model, you could probably assume that roll rate is proportional to control deflection.

Section 6.1.1.1-8 of the USAF DATCOM for stability and control about trailing edge plain flaps should answer your question in a very good way.

The rate of change of attitude also depends strongly on airspeed (local to that control surface;  in a tight turn the outboard aileron has more authority than the inboard one, for example).


text:
Identify odd aircraft near PWK 7/19/21
My workplace is under the pattern for Chicago Executive (PWK) and we sometimes get interesting/unusual things flying overhead.  Yesterday I heard something that sounded very different from a "normal" aircraft; I looked up and saw a straight-wing prop twin with canard, and what appeared (from below) to be just a thin pipe connecting the fuselage to the empennage (with  horizontal as well as vertical fin).  Maybe like a Piaggio / Diamond crossbreed?  I don't recall whether the props were in front or behind.  Sorry about the sparse description but that's the best I can do.
Any ideas?

Tractor (propellers in front of the wing) or pusher propellers?

It had both a canard and an aft horizontal tail?  Or does "empennage" just refer to the vertical fin?  Can't quite tell from your description.  Reference to "Piaggio" suggest the former but please clarify--

I've edited some more detail in.  I don't want to add too much since I fear when I try too hard my brain is adding details I may not have actually seen.  There was a full set of control surfaces aft.  I only got a glimpse for a few seconds.  The engine sound was quite distinctive though -- that's what made me look up.

Looking at more photos; it might have been a Piaggio.  From some angles the tail might look like an extension aft... https://www.wired.com/images_blogs/autopia/2010/08/Immagine-008-660x440.jpg

I wonder if this might have been an actual Diamond Hybrid -- a combustion/electric hybrid aircraft (hence strange engine sound -- the propellers are driven by electric motors).  The two electric motor nacelles (each holding an electric motor, batteries, and inverter) are mounted on a "canard" type surface (which doesn't appear to have any control surfaces, however) alongside the nose, and the very conventional empennage is on a fairly narrow tail cone.  I presume the combustion engine is inside the nose, as it would be on the conventional Diamond (single engine tractor).


text:
When a paraglider pilot applies brake on one side and maintains the brake pressure, why does the roll rate eventually come to zero?
I recently started to study paragliding and I am confused about the rolling movement created when a brake is maintained.
I was reading this answer How do paraglider controls work? and I am not clear which force should balance the rolling created by the side force on the levered arm during the initial phase of the turn and stop the rolling motion.

Is the question in essence simply "when a paraglider pilot applies brake on one side and maintains the brake pressure/ deflection indefinitely, why does the roll rate eventually come to zero?"

Thank you @quietflyer. Yes the question could be simplified with: "when a paraglider pilot applies brake on one side and maintains the brake pressure/ deflection indefinitely, why does the roll rate eventually come to zero?"
Yes I was talking about aerodynamic damping and yes it's not perfectly clear to me how it works. Thanks for the links. I am going to updated the question.

@Alessandro22-- you are welcome, the edits are a great improvement. I took down some of the comments but if you want to re-find some of the links google "Aviation Stack Exchange roll damping" -- but that's not really at the core of your question re paragliders I think-- "aerodynamic damping" relates to why the roll rate doesn't continue to INCREASE as we hold the ailerons, brakes, rudder, whatever, in a deflected position.  Aerodynamic damping doesn't actually pertain to an explanation of why continued control input is needed to maintain a given bank angle with ZERO roll rate.

In essence the question is "In a turn using brakes, when a paraglider pilot applies brake on one side and holds it indefinitely, why does the roll rate eventually come to zero?  Why must be substantial brake be applied just to maintain the bank angle in the turn?"
While turning using brakes1, to maintain the desired bank angle once it is established, paragliders do in fact typically require the pilot to maintain a substantial control input (brake pressure) in the same direction as was used to initiate the turn.  This is quite different from what we see in most conventional airplanes, and in modern hang gliders, but we do see similar dynamics in airplanes with lots of dihedral that lack ailerons and turn using the rudder or spoilers (examples-- "Gentle Lady" or "Radian" radio-controlled sailplane, Quicksilver MX Ultralight).  We also see similar dynamics in older hang gliders, where ample roll stability2 is created by  the highly swept shape of the delta or modified delta wing.  In many of these older hang gliders, additional factors contribute to roll stability.  These factors include a) some amount of dihedral, and b) a point of connection between the pilot's harness and the airframe that is well below the center of mass of the airframe, so that the resulting low CG contributes a strong stabilizing "pendulum effect" even when the pilot exerts zero force on the control bar.
For a basic description of paraglider turn dynamics, see this ASE answer -- How do paraglider controls work? . The answer explains how applying "brake" on one side modifies the lift vectors generated at various points along the arc of the wing in a way that generates a roll torque.
The present question is asking not what causes the roll torque into the turn when the brakes are applied, but rather why does the roll rate eventually drop to zero even with the brakes still applied. Clearly some aerodynamic force or torque is "trying" to roll the paraglider back toward wings-level.
The answer is that the turn involves some sideslip.  The aircraft is yawed to point slightly toward the outside or high side of the turn, relative to the direction of the flight path at any given instant. One cause of this sideslip is the simple fact that during turning flight, the outboard wingtip travels along a larger circumference and therefore must move through more air in a given unit time than the inboard wingtip, so it tends to experience more drag than the inboard wingtip.  The extra drag from the deployed brake on the inboard wingtip partly compensates for this, but not fully.  This makes the wing fly in an attitude where it is yawed to point slightly toward the outside of the turn relative to the actual direction of the flight path at any given instant.  In other words, the wing is constantly experiencing a sideways airflow component, toward the outside of the turn.
Notice that the arched shape of a paraglider wing exposes a tremendous amount of surface area to any sideways component in the airflow.  Ordinarily when we give an aircraft an anhedral geometry by lowering the wingtips in relation to the wing root, this interacts with any sideways airflow to generate a de-stabilizing roll torque, in the "upwind" direction, which in a slipping turn, would be away from wings-level.  This destabilizing roll torque component exists in the paraglider case too, but it is dwarfed by the stabilizing "downwind" roll torque caused by the fact that the center of area of the wing is so high above the CG of the whole system.  Since the wing is so high above the CG of the whole system, any aerodynamic sideforce generated by sideways airflow against the wing during the (slipping) turn will contribute a strong dihedral-like roll torque in the "downwind" direction, toward wings-level.  So will the wing's drag vector, because the glider is "pointing" a different direction than it is actually flying, and the drag vector acts parallel to the flight path.  Since these effects are both related to the fact that the CG of the whole system is located far below the center of area, they are sometimes collectively described as a "pendulum" effect, though we need to be cautious not the invoke the idea that the aircraft is somehow directly "feeling" the direction of the weight vector in a way that is not dependent on sideslip.3
So the answer to your question of why the bank angle does not continue to increase even when the brake application is maintained, is essentially this:
Turning flight in paragliders, as in many other rudderless aircraft, always involves some sideslip.
 The resulting sideways component in the relative wind acts on the center of area of the wing, high above the CG of the whole system, to generate a roll torque towards wings-level.
Here's one more nuance to consider in closing: we should note that when we initially apply "brake" in wings-level flight, the increased drag on the "braked" wing initially will "steer" (yaw) the glider into a skid in the intended direction of turn, so that the glider is actually pointing slightly toward the inside of the turn relative to actual flight path at any instant.  At this point the glider is "feeling" a sideways component in the relative wind toward the inside of the turn.  So at this point the interaction between the sideways airflow and the wing surface area, high above the CG, is actually generating a torque to help roll the wing toward a steeper bank angle.  Only once we establish a significant roll rate, and/ or
 a significant bank angle and turn rate, do we see the glider adopt an attitude where it is yawed slightly toward the outside of the turn, so that is "slipping" rather than "skidding".  (This answer has focused on how turning tends to cause sideslip; to read about how rolling also tends to cause sideslip see this section on adverse yaw from the excellent "See How It Flies" website.)  All the same can be said of an airplane with lots of dihedral that has no ailerons but rather uses the rudder for roll control-- like many radio-controlled sailplanes-- when we first apply rudder to enter a turn, we know the aircraft must be skidding, but as the bank angle increases and we settle into a steady turn at constant bank angle, we know the aircraft must in fact be slipping, even though we are continuing to hold some rudder input into the turn.  If the aircraft were not slipping at this point, the net roll torque could not be zero, due to the fact that the wing on the outside of the turn is moving faster through the air, and tending to generating more lift, than the wing on the inside of the turn.
Highly related -- Does "pendulum effect" apply to hang gliders or any aircraft?
Footnotes --
1 -- It is not the intent of this answer to suggest that a turn in paragliders is always, or typically, or ideally, accomplished using brakes alone with no weight-shift.  Weight-shift, where the pilot leans in the harness to rock his body to one side and load up one side of the suspension system and unload the other side, is another way to turn.  Both methods may be used in combination.  However the original question was focused on a turn carried out with the brakes, so this answer is as well.
2 -- This answer uses "roll stability" to mean a tendency to roll toward wings-level.  Other alternative terms might be used to avoid suggesting that the aircraft is tending to maintain a given bank angle, but "roll stability" is the phrase we'll stick with for this answer.
3 -- "Keel effect" is another term that is sometimes used to describe the stabilizing effect of a low CG position, though this is also somewhat problematic, for two reasons.  A) It would be better to call it the "keel weight effect", because an unweighted keel (think of a lightweight centerboard that may be raised or lowered at will) actually contributes a net hydrodynamic roll torque in the opposite direction during sideslip (sideways drift), tending to roll the boat in the upstream direction.  B) The righting effect of a heavy keel weight is actually not dependent on sideslip at all-- buoyancy creates a righting moment even when total velocity, and therefore sideslip velocity, is zero-- so we still have the issue that we were trying to avoid by not using the term "pendulum effect", namely that we'd like to avoid suggesting that a (non-buoyant, e.g. heavier-than-air) aircraft in flight can somehow "feel" the direction of the gravity vector and therefore "sense" when the aircraft is tilted away from level, even if sideslip is somehow completely eliminated by appropriate rudder usage or by any other means.
Addendum--
This answer has referenced a "skid" as the turn is first initiated with brake, and a "slip" once the turn is established.  This begs the question, "what about the roll torque created as the pilot's body tends to swing to the left or the right, just as a slip-skid (inclinometer) ball would, due to the apparent forces created by the slip or the skid?  Doesn't this play a key role in the balance of roll torques on a paraglider?"
Or in other words, "Isn't there some additional sort of 'pendulum effect' that we have not yet fully considered?"
At least one answer to the related question "How do paraglider controls work?" has in fact focused on apparent inertial forces caused by the pilot's body "not want[ing] to turn (yet)", purportedly causing the pilot's body to "swing" against the intended direction of turn as the turn is being initiated, thus creating a roll torque.
The truth is that we can consider the paraglider and pilot to be a single, essentially rigid sytem.  The lines do not go slack in normal flight.  The pilot is fixed in position under the wing by the triangular geometry of the multiple suspension lines.  If we want to analyze the motion of the whole body, it is not necessary to take into account the fictitious "centrifugal force" created by turning, or any other apparent inertial force.  (An exception would arise if the aircraft were rotating very rapidly about one of its axes-- for example a very high rate of yaw rotation, somewhat similar to a flat spin in a conventional aircraft, might tend to flatten out the canopy into more of a horizontal line, due to "centrifugal force".  Similarly, very high rates of pitch rotation play a key role in the dynamics of certain aerobatic maneuvers, such as spiral dives, in paragliders.)
The apparent tendency of the pilot to "swing" toward the inside of the turn in a slip, and toward the "outside" of the turn in a skid, thus imparting a roll torque to the wing, is in fact already fully accounted for in this answer.  We account for that roll torque when we note that the sideforce created by the impact of the sideways airflow against wing acts high above the CG of the whole system and thus creates a roll torque.  The apparent sideways force experienced in the reference frame of the pilot in a slip or skid, is in fact simply the mirror image of the sideways component of the real aerodynamic force generated by the aircraft during the slip or skid.  If the aircraft could somehow slip or skid sideways through the air without generating any aerodynamic sideforce, then no roll torque would result.  Also, if the aircraft could somehow slip or skid sideways through the air without generating any aerodynamic sideforce, then the pilot would experience no apparent tendency to swing to one side.
After noting that during a slip or skid, the sideways airflow against the wing generates an aerodynamic sideforce that acts high above the CG of the whole aircraft-pilot system and thus generates a roll torque, it would be erroneous to suggest that the apparent tendency of the pilot's body to "swing to one side" in a slip or a skid somehow contributes an additional roll torque about the wing's center of lift or about any other point.
Additional content on ASE relating to the ideas in this addendum--
(Q) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?
(A) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?
(A) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?
(A) Forces in a slipping turn

A recent edit suggested by another ASE participant introduced "white space" by breaking up some of the paragraphs.  It seemed to me to make it less clear which ideas were most closely related to which other ideas, and made the "flow" of the argument harder to understand.  I decided to reject it.  If someone would like to explain to me how this edit was an improvement I'd be happy to listen, but I'm not seeing it at present.  Fixing the direction of the quotation marks was fine, but I don't understand the reason for breaking up the paragraphs.

RE "we can consider the paraglider and pilot to be a single, essentially rigid sytem" - A friend pointed out that this is not true in the vertical axis. When you apply a brake input and the wing starts to yaw, there is a twist about the carabineers that is only resisted by the fabric connecting the two "halves" of the wing.

@OlegPryadko -- thanks for the note, need to take some time to think about that.


text:
What is SVC CTC on a chart supplement?
The chart supplement for KSNA mentions the following

AIRSPACE: CLASS C svc ctc APP CON svc 1415–0700Z‡ other times CLASS G.

I know that SVC means service. I've gathered that KSNA is class C from 1415 - 0700Z (which is 6:15am - 11pm PDT local time). My questions are the following.

It says "svc ctc" and "APP CON svc." What does this mean? My first guess is, both control tower and approach control are in effect. Is that right?
What is ctc? Control tower?

FAA Aeronautical Chart Users' Guide web page has a link to pdf guide (22MB), lists the abbreviations as follows:

svc = service
ctc = contact
APP = approach control

abbreviation con is not listed.
Aeronautical Chart Users' Guide web page (no, this is not the same page as above) provides link to a different pdf (direct link to full guide, 52MB) that provides explanation to abbreviation

APP CON = approach control

So "svc ctc" would decipher as "for service contact" and latter svc followed by timeframe would inform of applicable service hours of the ATC entity. The whole part in question would thus read "AIRSPACE CLASS C service contact APPROACH CONTROL, service available 1415-0700 UTC (adjusted for DST); other times CLASS G".

Oops, not my area of expertise really... mega something, quite big... I'll fix that in a bit, thanks!

Related: https://aviation.stackexchange.com/questions/91022/which-frequency-to-contact-when-arriving-at-a-class-c-airport-when-the-tower-is -- the present answer, which I suspect is correct, supports the idea that there's no need to contact Approach Control when the airspace is Class G at the surface.


text:
Could you help me identify this plane (miniature model)?

please help me identify this model.
Thank you

Both the Sopwith Camel and the Nieuport 11 mentioned elsewhere had rotary engines with a rounded cowling, but the model in the photograph appears to have a square cowling. This suggests it might be a Royal Aircraft Factory S.E.5 or S.E.5a which had a V8 engine.

Wikimedia Commons

The model in the original post is a British Royal Aircraft Factory S.E.5A - the shape is quite distinctive, and it was a popular subject for balsa modellers due to its boxy fuselage being simple to build.
Here is a drawing of it.

This particular model appears to have ailerons on the lower wing only, and they are much wider in span and narrower in chord than the ailerons typical of aircraft of this era.  This strongly suggests that this is a radio-controlled flying model, and that extreme scale accuracy was not of utmost importance to the builder.  This observation may bear strongly on whether or not potential candidates should be eliminated based on various small (or large) differences from the model.  The SE-5a in particular comes to mind, and differs from the actual model in the photo in the question in all the following areas--

Model has wide-span, narrow chord, ailerons, on lower wing only, as noted above.

Model lacks an axle connecting the two wheels.

On the model, a close look at the geometry of the apparent visual gap (i.e. as seen from that particular viewing angle) between the TE of the bottom wing and the LE of the top wing shows a wedged-shape gap that indicates a difference in sweep (more on the top wing), or a difference in dihedral angle (more on the top wing), or both.

Based on point #1, it seems reasonable to suggest that points 2 and 3 may simply have resulted from the model builder's lack of interest in extreme accuracy, and that the aircraft may well be a representation of an SE-5a.  (If so, the wings of the model are undoubtedly all unswept, like the original, and point #3 arises from more dihedral in the model in the top wing than the bottom wing.)
Point #2 in particular may be simply an accommodation to the practicalities of operating a model plane off grass fields or other "rough" surfaces.

excellent observations! +1

I believe it might be a Sopwith Camel, though I’m no expert.

Could be a Nieuport 11 too. Hard to say without seeing more of the tail.


text:
Class G Airspace ceiling: 700' versus 1200'
Why does on various images, such as below, have 1200' and 700' AGL for Class G Airspace?
Why is it sometimes 1200' and other times 700' AGL?
Other than the altitude, how are the 1200' and 700' different? And why these two altitudes are important for Class G airspace.
https://www.aerial-guide.com/article/sectional-chart-airspace-classification-overview

It’s the height at which the airspace changes from class G to class E.  Does that answer your question or is there more that you want to know?

Why is it sometimes 1200' and other times 700' AGL?

It’s uncontrolled airspace, so it may stop at a height where controlled airspace starts, or for whatever reason was chosen by whoever planned that particular airspace.  There isn’t one rule for all blocks of airspace, and nothing special about 700 and 1200ft

This answer will be aimed at the US.

Why does on various images, such as below, have 1200' and 700' AGL for
Class G Airspace?

The "1200' AGL" and "700' AGL" labels are simply stating the altitude where the figure is intending to show that the Class G airspace ends and the Class E airspace begins, at different locations.
Note that it is very common one for of these two altitudes to be used for the floor of Class E airspace (and therefore the ceiling of the underlying Class G airspace), and a specific symbol is dedicated to each of these two cases on VFR sectional charts.
But if--as you suggested in comments-- you were asking what criteria is used for deciding which of these altitudes should be used for the floor of any particular Class E airspace (and therefore the ceiling of the underlying Class G airspace), read on--
In the US, airspace designers follow the guidance in a FAA document called "Procedures for Handling Airspace Matters" (see link here).
On page 18-2-1, we read the following:

18−2−2. 700/1,200−FOOT CLASS E AIRSPACE
Class E−5 700/1200−foot airspace areas are used for transitioning
aircraft to/from the terminal or en route environment.
18−2−3. 700−FOOT CLASS E AIRSPACE
A Class E−5 airspace area with a base of 700 feet above the surface
must be designated to accommodate arriving IFR operations below 1,500
feet above the surface and departing IFR operations until they reach
1,200 feet above the surface.
18−2−4. 1,200−FOOT CLASS E AIRSPACE
Where sufficient controlled airspace does not exist, designate a 1,200
foot Class E−5 airspace area to accommodate arriving IFR operations at
1,500 feet and higher above the surface and departing IFR operations
from the point they reach 1,200 feet above the surface until reaching
overlying or adjacent controlled airspace.
18−2−5. CLASS E AIRSPACE FLOORS ABOVE 1,200 FEET
Class E−5 airspace areas may be established with MSL floors above
1,200 feet AGL. Normally floors will be at least 300 feet below the
minimum IFR altitude.

So, a Class E floor of 700' AGL rather than 1200' AGL will be set in areas where arriving IFR traffic is expected to be operating below 1500' AGL or departing IFR traffic is expected to be operating below 1200' AGL, unless the airspace is protected by some higher level of protection, such as Class B, C, or D airspace, or Class E airspace extending all the way down to the surface.
Sections 18-3-1 through 18-3-6 go into much more detail on how these Class E floors should be set to match the expected paths of departing and arriving IFR traffic.
Interestingly, the "Procedures for Handling Airspace Matters" document is rather vague on what criteria should be used to designate areas where the floor of Class E airspace comes all the way down to the surface.  The answer lies in part in section 17-2-7 (d) on page 17-2-2, but this only addresses surface-level Class E "extensions" to Class D or higher airspace, not situations like at KSHR or KONP, where a circle (or other shape) of surface-level Class E airspace has been designated to protect an airport that entirely lacks any Class D (or higher) airspace.  The latter situation is addressed in sections 18-1-2 and 18-1-3 on pages 18-1-1 and 18-1-2, but the criteria given are much less specific than the ones that were given for setting Class E floors at 700' AGL or 1200' AGL.